The thermal decomposition of a basic carbonate of a certain main group element is as follows: Accurately weigh $1.000\ \mathrm {g}$ of the salt, heat to about $300\mathrm { ℃ }$ to undergo the first step of decomposition with a weight loss, producing two gases with a total weight of $0.1712\mathrm { g }$, cool to room temperature, and the remaining gaseous product is only $0.0941\mathrm { g }$. Continue to heat the remaining solid product, and the second step of decomposition with a weight loss occurs at about $425\mathrm {℃}$, and the mass of the remaining solid product is $0.5721{ \mathrm { g } }$. Continue heating to constant weight, and the mass of the final product is measured to be $0.4310{ \mathrm { g } }$. There are the following statements:

1. The element is +1 valence

2. The mass fraction of the metal in the basic carbonate is $0.2599$

3. The final solid product remaining after heating is easily soluble in water

4. Obtain all the gases released during the above heating process, cool them to room temperature, and the mass of the remaining gas is $0.4919{ \mathrm { g } }$

5. The number of gas molecules released in the second step of weight loss is 3 times the number of gas molecules released upon heating to constant weight; the number of gas molecules released in the first step of weight loss is 2 times the number of gas molecules released upon heating to constant weight.

6. The solid product of the second step of weight loss contains 3 elements

Select the option below that contains the most correct options
某主族元素的碱式碳酸盐热分解情况如下：准确称取 $1.000\ \mathrm {g}$ 该盐，加热至 $300\mathrm { ℃ }$ 左右发生第一步分解失重产生两种气体共重 $0.1712\mathrm { g }$ ，冷却至室温，气态产物仅剩 $0.0941\mathrm { g }$ 。继续加热剩余固态产物，在 $425\mathrm {℃}$ 左右发生第二步分解失重，剩余固态产物的质量为$0.5721{ \mathrm { g } }$。继续加热至恒重，测得最终产物的质量为 $0.4310{ \mathrm { g } }$ 。有以下说法：

1.该元素为+1价

2.该碱式碳酸盐中金属的质量分数为$0.2599$

3.加热剩余的最终固态产物易溶于水

4.获取以上加热过程中释放的所有气体，将其冷却至室温后，剩余气体质量为$0.4919{ \mathrm { g } }$

5.第二步失重释放的气体分子数是加热至恒重释放的气体分子数的3倍；第一步失重释放的气体分子数是加热至恒重释放的气体分子数的2倍

6.第二步失重的固体产物中含3种元素

选出下列选项中含正确选项最多的一项
1,2,3
2,3,5
1,3,5
2,5,6
1,5,6
3,4,6
4,5,6
2,4,6
1,4,5
1,2,4
2,3,4
1,2,3
2,3,5
1,3,5
2,5,6
1,5,6
3,4,6
4,5,6
2,4,6
1,4,5
1,2,4
2,3,4
Cooled to room temperature, the mass of the gaseous product decreased, obviously due to $\mathrm {H_2O}$. Then the generated water vapor $\mathrm { m }(\mathrm {H_2O}) = 0.1712 - 0.0941 = 0.0771 \mathrm { g }$. The remaining gas from the thermal decomposition of the basic carbonate is $\mathrm {CO_ 2}$ , then $\mathrm { n }(\mathrm {CO_2}) : \mathrm { n }(\mathrm { H_2O })= 0.0941/44.01 : 0.0771/18.016 = 0.002138 : 0.004280 = 1 : 2$

<Checkpoint>[The portion of the gaseous product mass reduction upon cooling to room temperature is $\mathrm { H_2O}$](1)</Checkpoint>

<Checkpoint>[Calculated, the ratio of the number of molecules released of $\mathrm { CO_2 }$ to $\mathrm {H_2O}$ is 1:2](1)</Checkpoint>

The second step weight loss $(1.000-0.5721)-0.1712 = 0.2567 \mathrm { g }$, based on the amount of substance of $\mathrm { CO_2 }$ released in the first step weight loss, the corresponding relative molecular mass of the gas is $0.2567 / 0.002138 = 120.02$, corresponding to 3 $\mathrm {H_2O}$ and 1.5 $\mathrm { C O_ 2 }$

<Checkpoint>[Calculated, the second step weight loss produces 3 $\mathrm { H_2O }$ and 1.5 $\mathrm { CO_2 }$](1)</Checkpoint>

Similarly, the third step weight loss $0.5721 - 0.4310 = 0.1411 \mathrm { g }$, corresponding to 1.5 $\mathrm { CO_2 }$, then the molecular weight of the gas produced in the first step weight loss is twice that produced by heating to constant weight; then the molecular weight of the gas produced in the second step weight loss is 3 times that produced by heating to constant weight, statement 5 is correct.

<Checkpoint>[Calculated, the third step weight loss produces 1.5 $\mathrm { CO_2 }$](1)</Checkpoint>

The final product should be an oxide. Let the molecular weight of the oxide be x. A total of 4 $\mathrm { CO_2 }$ and 5 $\mathrm { H_2O }$ are lost. According to the question, the equation is $\mathrm {nx / (nx + 4× 44.01 + 5 × 18.016) = 0.4310 / 1.000 }$. Simplifying, we get $\mathrm {x = 201.6 / n}$. When $n = 5$, there is a reasonable solution, corresponding to the oxide $\mathrm { MgO }$. $\mathrm { MgO }$ is not easily soluble in water, statement 3 is incorrect.

<Checkpoint>[Calculated, the oxide is $\mathrm { MgO }$, $\mathrm { MgO }$ is not easily soluble in water](1)</Checkpoint>

Therefore, the chemical formula of the substance is $\mathrm {5MgO·4CO_2·5H_2O}$, the magnesium element content is $0.2599$ and it is +2 valent, statement 2 is correct, statement 1 is incorrect.

<Checkpoint>[Inferred that the chemical formula of the basic carbonate is $\mathrm {5MgO·4CO_2·5H_2O}$, containing metal $0.2599$ and it is +2 valent](1)</Checkpoint>

A total of 2.5 $\mathrm { CO_2 }$ and 5 $\mathrm { H_2O }$ were released after the second step weight loss. The remaining solid composition is $\mathrm {5MgO·1.5CO_2}$, containing three elements, statement 6 is correct.

<Checkpoint>[After the second step weight loss, all $\mathrm { H_2O }$ is released, and the remaining solid composition is $\mathrm {5MgO·1.5CO_2}$, containing three elements](1)</Checkpoint>

After the released gas returns to room temperature, only $\mathrm { CO_2 }$ can remain gaseous. A total of $1.000 / [5 × (24. 31 + 16) + 4 × 44.01 + 5 × 18.016] × 4 × 44.01 = 0.3764 \mathrm {g}$ can be released, statement 4 is incorrect.

<Checkpoint>[The released $\mathrm { CO_2 }$ remains gaseous after returning to room temperature, with a total mass of $0.3764 \mathrm {g}$](1)</Checkpoint>

Select option D
冷却至室温气态产物质量减少，显然为 $\mathrm {H_2O}$。则生成水蒸气 $\mathrm { m }(\mathrm {H_2O}) = 0.1712 - 0.0941 = 0.0771 \mathrm { g }$。碱式碳酸盐热分解剩余气体为 $\mathrm {CO_ 2}$ ，则$\mathrm { n }(\mathrm {CO_2}) : \mathrm { n }(\mathrm { H_2O })= 0.0941/44.01 : 0.0771/18.016 = 0.002138 : 0.004280 = 1 : 2$ 

<Checkpoint>[冷却至室温气态产物质量减少的部分为 $\mathrm { H_2O}$](1)</Checkpoint>

<Checkpoint>[计算得释放出$\mathrm { CO_2 }$与$\mathrm {H_2O}$的分子数之比为1：2](1)</Checkpoint>

第二步失重 $(1.000-0.5721)-0.1712 = 0.2567 \mathrm { g }$ ，以第一步失重释放$\mathrm { CO_2 }$的物质的量为基准，对应气体的相对分子质量为$0.2567 / 0.002138 = 120.02$，对应3个$\mathrm {H_2O}$和1.5个$\mathrm { C O_ 2 }$

<Checkpoint>[计算得第二步失重产生3个$\mathrm { H_2O }$和1.5个$\mathrm { CO_2 }$](1)</Checkpoint>

同理，第三步失重 $0.5721 - 0.4310 = 0.1411 \mathrm { g }$，对应1.5个 $\mathrm { CO_2 }$，则第一步失重产生的气体分子量是加热至恒重产生的2倍；则第二步失重产生的气体分子量是加热至恒重产生的3倍，说法5正确

<Checkpoint>[计算得第三步失重产生1.5个$\mathrm { CO_2 }$](1)</Checkpoint>

最终产物应为氧化物，设该氧化物的分子量为x，总共失去了4个 $\mathrm { CO_2 }$和5个 $\mathrm { H_2O }$，根据题意列方程得 $\mathrm {nx / (nx + 4× 44.01 + 5 × 18.016) = 0.4310 / 1.000 }$，化简得，$\mathrm {x = 201.6 / n}$，当 $n = 5$ 时有合理解，对应氧化物$\mathrm { MgO }$，$\mathrm { MgO }$不易溶于水，说法3错误。

<Checkpoint>[计算得氧化物为$\mathrm { MgO }$，$\mathrm { MgO }$不易溶于水](1)</Checkpoint>

因此该物质的化学式为$\mathrm {5MgO·4CO_2·5H_2O}$，镁元素含量为$0.2599$且为+2价，说法2正确，说法1错误。

<Checkpoint>[推断出该碱式碳酸盐的化学式为$\mathrm {5MgO·4CO_2·5H_2O}$，其中含金属$0.2599$且为+2价](1)</Checkpoint>

第二步失重后一共释放了2.5个$\mathrm { CO_2 }$和5个$\mathrm { H_2O }$，剩余固体组成为$\mathrm {5MgO·1.5CO_2}$，含三种元素，说法6正确。

<Checkpoint>[第二步失重后释放了全部$\mathrm { H_2O }$，剩余固体组成为$\mathrm {5MgO·1.5CO_2}$，含三种元素](1)</Checkpoint>

释放的气体在恢复室温后，仅有$\mathrm { CO_2 }$可继续保持气态，共可释放出$1.000 / [5 × (24. 31 + 16) + 4 × 44.01 + 5 × 18.016] × 4 × 44.01 = 0.3764 \mathrm {g}$，说法4错误。 

<Checkpoint>[释放的$\mathrm { CO_2 }$在恢复到室温后仍保持气态，总质量为$0.3764 \mathrm {g}$](1)</Checkpoint>

选择选项D